Jordan normal form of powers of Jordan normal form Previous related question: Jordan normal form powers

Let $A$ be a $n\times n$ Matrix such that $A=PBP^{-1}$ where $B$ is in Jordan normal form with $\lambda_i(k)_j$ Where $i$ is the size, $k$ is the eigenvalue and $j$ the order.

From the previous question I know that each Jordan block $\lambda_i(k)_j$ when the matrix is raised to the $n$-th power is an upper triangular matrix $$\sum_{r=0}^{i-1} {n \choose r} k^{n-r}t^r$$
Where $t$ is the matrix with 1’s on it’s super diagonal and 0’s everywhere else.
How can I get this matrix to Jordan normal form?
 A: So you want to know the Jordan canonical form of the $i \times i$ matrix
$$ A = \sum_{r=0}^{i-1} \left( n \atop r \right) k^{n-r} t^r .$$
Since $A$ has $k^n$ as an $i$-fold repeated eigenvalue, it is sufficient to find the Jordan form for
$$ A - k^n I = \sum_{r=1}^{i-1} \left( n \atop r \right) k^{n-r} t^r .$$
First consider the case $k \ne 0$.  Then
$$ (A- k^n I)^{i-1} = n^{i-1} k^{(n-1)(i-1)} t^{i-1} \ne 0$$
since $t^r = 0$ for $r \ge i$.  Similarly $(A- k^n I)^i = 0$.  Therefore the minimal polynomial for $A$ is $p(x) = (x - k^n)^i$, and its Jordan canonical form must be $k^n I + t$, that is, a single block of size $i$.
Next, consider the case $k = 0$, when $A = t^n$.  Denote the unit vectors by $e_r$ with $1 \le r \le i$.  Then the unit vectors split into groups:

*

*$e_1, e_{n+1}, e_{2n+1}, \dots$ of size $[(i+n-1)/n]$;

*$e_2, e_{n+2}, e_{2n+2}, \dots$ of size $[(i+n-2)/n]$;

*$e_3, e_{n+3}, e_{2n+3}, \dots$ of size $[(i+n-3)/n]$;

*$\vdots$

*$e_n, e_{2n}, e_{3n}, \dots$ of size $[i/n]$;

where $[x]$ denotes the integer part of $x$.  On each group, $A$ acts as a Jordan block.  So its Jordan canonical form is a collection of blocks of size $[(i+n-1)/n], [(i+n-2)/n], \dots, [i/n]$.  And if you think about it, this is $n - i + n[i/n]$ blocks of size $[i/n]$ and $i - n[i/n]$ blocks of size $[i/n]+1$.  (In particular, if $n \ge i$, then it is $i$ blocks of size $1$, that is, $A = 0$ is diagonal.)
